Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Definite Integration question

2023 · 13 Apr · Shift 1 · Q30
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Definite Integration
  5. /2023 · 13 Apr · Shift 1 · Q30

Definite Integration question

2023 · 13 Apr · Shift 1 · Q30

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
∫0∞6e3x+6e2x+11ex+6dx=\int_{0}^{\infty} \frac{6}{e^{3 x}+6 e^{2 x}+11 e^{x}+6} d x=∫0∞​e3x+6e2x+11ex+66​dx=
  1. A
    log⁡e(25681)\log _{e}\left(\frac{256}{81}\right)loge​(81256​)
  2. B
    log⁡e(6427)\log _{e}\left(\frac{64}{27}\right)loge​(2764​)
  3. C
    log⁡e(3227)\log _{e}\left(\frac{32}{27}\right)loge​(2732​)
  4. D
    log⁡e(51281)\log _{e}\left(\frac{512}{81}\right)loge​(81512​)
View written solutionFree

Correct answer: C

  1. Evaluate the integral

We need to compute I=∫0∞6e3x+6e2x+11ex+6 dx.I=\int_{0}^{\infty} \frac{6}{e^{3x}+6e^{2x}+11e^x+6}\,dx.I=∫0∞​e3x+6e2x+11ex+66​dx.

  1. Substitute t=ext=e^xt=ex

Since t=ext=e^xt=ex, we have dt=exdx=t dx  ⟹  dx=dtt.dt=e^x dx=t\,dx \implies dx=\frac{dt}{t}.dt=exdx=tdx⟹dx=tdt​.

Also, when x=0x=0x=0, t=1t=1t=1; and when x→∞x\to\inftyx→∞, t→∞t\to\inftyt→∞.

So, I=∫1∞6t3+6t2+11t+6⋅dttI=\int_{1}^{\infty} \frac{6}{t^3+6t^2+11t+6}\cdot \frac{dt}{t}I=∫1∞​t3+6t2+11t+66​⋅tdt​ =∫1∞6t(t3+6t2+11t+6) dt.=\int_{1}^{\infty} \frac{6}{t(t^3+6t^2+11t+6)}\,dt.=∫1∞​t(t3+6t2+11t+6)6​dt.

Now factor the cubic: t3+6t2+11t+6=(t+1)(t+2)(t+3).t^3+6t^2+11t+6=(t+1)(t+2)(t+3).t3+6t2+11t+6=(t+1)(t+2)(t+3).

Hence, I=∫1∞6t(t+1)(t+2)(t+3) dt.I=\int_{1}^{\infty} \frac{6}{t(t+1)(t+2)(t+3)}\,dt.I=∫1∞​t(t+1)(t+2)(t+3)6​dt.

  1. Use partial fractions

Write 6t(t+1)(t+2)(t+3)=At+Bt+1+Ct+2+Dt+3.\frac{6}{t(t+1)(t+2)(t+3)}=\frac{A}{t}+\frac{B}{t+1}+\frac{C}{t+2}+\frac{D}{t+3}.t(t+1)(t+2)(t+3)6​=tA​+t+1B​+t+2C​+t+3D​.

Using standard cover-up values:

  • At t=0t=0t=0: 6=A(1)(2)(3)  ⟹  A=1.6=A(1)(2)(3) \implies A=1.6=A(1)(2)(3)⟹A=1.

  • At t=−1t=-1t=−1: 6=B(−1)(1)(2)  ⟹  B=−3.6=B(-1)(1)(2) \implies B=-3.6=B(−1)(1)(2)⟹B=−3.

  • At t=−2t=-2t=−2: 6=C(−2)(−1)(1)  ⟹  C=3.6=C(-2)(-1)(1) \implies C=3.6=C(−2)(−1)(1)⟹C=3.

  • At t=−3t=-3t=−3: 6=D(−3)(−2)(−1)  ⟹  D=−1.6=D(-3)(-2)(-1) \implies D=-1.6=D(−3)(−2)(−1)⟹D=−1.

Therefore, 6t(t+1)(t+2)(t+3)=1t−3t+1+3t+2−1t+3.\frac{6}{t(t+1)(t+2)(t+3)}=\frac{1}{t}-\frac{3}{t+1}+\frac{3}{t+2}-\frac{1}{t+3}.t(t+1)(t+2)(t+3)6​=t1​−t+13​+t+23​−t+31​.

  1. Integrate term by term

So, I=∫1∞(1t−3t+1+3t+2−1t+3)dt.I=\int_1^{\infty}\left(\frac1t-\frac3{t+1}+\frac3{t+2}-\frac1{t+3}\right)dt.I=∫1∞​(t1​−t+13​+t+23​−t+31​)dt.

An antiderivative is ln⁡t−3ln⁡(t+1)+3ln⁡(t+2)−ln⁡(t+3).\ln t-3\ln(t+1)+3\ln(t+2)-\ln(t+3).lnt−3ln(t+1)+3ln(t+2)−ln(t+3).

Thus, I=[ln⁡t−3ln⁡(t+1)+3ln⁡(t+2)−ln⁡(t+3)]1∞.I=\left[\ln t-3\ln(t+1)+3\ln(t+2)-\ln(t+3)\right]_1^{\infty}.I=[lnt−3ln(t+1)+3ln(t+2)−ln(t+3)]1∞​.

  1. Evaluate the limit at infinity

Combine the logarithms: ln⁡(t(t+2)3(t+1)3(t+3)).\ln\left(\frac{t(t+2)^3}{(t+1)^3(t+3)}\right).ln((t+1)3(t+3)t(t+2)3​).

As t→∞t\to\inftyt→∞, t(t+2)3(t+1)3(t+3)→1,\frac{t(t+2)^3}{(t+1)^3(t+3)}\to 1,(t+1)3(t+3)t(t+2)3​→1, so the upper limit is ln⁡1=0.\ln 1=0.ln1=0.

  1. Evaluate at t=1t=1t=1

At t=1t=1t=1, ln⁡1−3ln⁡2+3ln⁡3−ln⁡4\ln 1-3\ln 2+3\ln 3-\ln 4ln1−3ln2+3ln3−ln4 =−3ln⁡2+3ln⁡3−2ln⁡2= -3\ln 2+3\ln 3-2\ln 2=−3ln2+3ln3−2ln2 =3ln⁡3−5ln⁡2.=3\ln 3-5\ln 2.=3ln3−5ln2.

Therefore, I=0−(3ln⁡3−5ln⁡2)=5ln⁡2−3ln⁡3.I=0-(3\ln 3-5\ln 2)=5\ln 2-3\ln 3.I=0−(3ln3−5ln2)=5ln2−3ln3.

So, I=ln⁡(2533)=ln⁡(3227).I=\ln\left(\frac{2^5}{3^3}\right)=\ln\left(\frac{32}{27}\right).I=ln(3325​)=ln(2732​).

  1. Match with the options

ln⁡(3227)\ln\left(\frac{32}{27}\right)ln(2732​) corresponds to Option C.

PreviousNext

More from Definite Integration

  • Let for x∈R,S0​(x)=x,Sk​(x)=Ck​x+k∫0x​Sk−1​(t)dt, where C0​=1,Ck​=1−∫01​Sk−1​(x)dx,k=1,2,3,… Then S2​(3)+6C3​ is equal to ​.2023 · Numerical
  • The value of 0∫4π​​e−x(tan49x+tan51x)dxe−4π​+0∫4π​​e−xtan50xdx​ is2023 · MCQ
  • Let fn​=∫02π​​(∑k=1n​sink−1x)(∑k=1n​(2k−1)sink−1x)cosxdx,n∈N. Then f21​−f20​ is equal to ​2023 · Numerical
  • If 0∫1​(5+2x−2x2)(1+e(2−4x))1​dx=α1​loge​(βα+1​),α,β>0, then α4−β4 is equal to :2023 · MCQ
  • The value of 120∫3​​x2−3x+2​dx is ​2023 · Numerical
  • The value of π8​0∫2π​​(sinx)2023+(cosx)2023(cosx)2023​dx is ​2023 · Numerical
  • 432​​∫433​​​9−4x2​48​dx is equal to :2023 · MCQ
  • Let f be a differentiable function defined on [0,2π​] such that f(x)>0 and f(x)+∫0x​f(t)1−(loge​f(t))2​dt=e,∀x∈[0,2π​]. Then (6loge​f(6π​))2…2023 · Numerical